# Is this a correct way to prove uniqueness using limits?

I have a question about the following proof:

Claim: A sequence in $\mathbb{R}$ can have at most one limit.

Proof: Assume a sequence $X = (x_n)$ has two limits. Call them $x$ and
$x’$. For any $\epsilon > 0$, there exists $N$ such that $|x_n – x| < \epsilon/2$ for $n \geq N$. There also exists $N’$ such that $|x_{n’}- x’| < \epsilon/2$ for $n’ \geq N’$. Let $M =$max$(N, N’)$. Then for $m \geq M$: $|x-x’| = |(x – x_m) + (x_m – x’)| \leq |x_m – x| + |x_m – x’| < \epsilon/2 + \epsilon/2= \epsilon$.

Since $\epsilon$ is arbitrary, we conclude $x = x’$.

My question: is it necessary to split $\epsilon$ at all? Do we do this just because we want a clean looking proof? Here is the alternative proof which I had in mind, which may or may not be correct:

Assume a sequence $X = (x_n)$ has two limits. Call them $x$ and
$x’$. For any $\epsilon > 0$, there exists $N$ such that $|x_n – x| < \epsilon$ for $n \geq N$. For $\epsilon_2 > 0$, there exists $N’$ such that $|x_{n’}- x’| < \epsilon_2$ for $n’ \geq N’$. Let $M =$max$(N, N’)$. Then for $m \geq M$: $|x-x’| = |(x – x_m) + (x_m – x’)| \leq |x_m – x| + |x_m -x’| < \epsilon + \epsilon_2= \epsilon_3$.

$\epsilon_3$ is just another positive real number. I can make $\epsilon_3$ as small as I like because I can make $\epsilon$ and $\epsilon_2$ as small as I like. So I draw the same conclusion.

Is the last bit of reasoning valid? I’ve gotten the impression, from speaking with a professor, that I introduce a dependency on $\epsilon$ and $\epsilon_2$, so I should be able to come up with some method of getting $\epsilon$ and $\epsilon_2$. I’m confused.

(the angle brackets from the blockquote seem to be screwing up the formatting.)

#### Solutions Collecting From Web of "Is this a correct way to prove uniqueness using limits?"

Both proofs are correct. Your statement that you can make $\epsilon_1 + \epsilon_2$ as small as you like (at the end of the second proof) is of course correct, but could be made more precise by saying something like “given any $\epsilon > 0$, if we set $\epsilon_1,\epsilon_2 = \epsilon/2$ then we have $\epsilon_1 + \epsilon_2 \le \epsilon$.” Then you would be saying the same thing as the first proof.